Question 24.10: Use the properties of the delta function to deduce its Fouri......

By definition

{\mathcal{F}}\{\delta(t-a)\}=\int_{-\infty}^{\infty}\delta(t-a)\,\mathrm{e}^{-\mathrm{j}\omega t}\,\mathrm{d}t

Next, recall the following property of the delta function

\int_{-\infty}^{\infty}f(t)\delta(t-a)\,\mathrm{d}t=f(a)                               (24.9)

for any reasonably well-behaved function f (t). Using Equation (24.9) with f (t ) = e^−{j \omega t}  we have

In particular, if a = 0 we have {\mathcal{F}}\{\delta(t)\}=1. This result is depicted in Figure 24.9.